437. Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11

题解

此题的难点需想清每个节点都要当作根节点来对待, 因此题解为如下三部分之和:

  1. 当前节点作为根节点时相加为 sum 的路径之和。
  2. 当前节点左子树作为根节点时相加为 sum 的路径之和。
  3. 当前节点右子树作为根节点时相加为 sum 的路径之和。

同时递归可以分为两个部分: 根节点的递归分析总和的递归

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number}
*/
var pathSum = function(root, sum) {
if (!root) return 0
// total means result value
// initialSum means init sum value
let result = { total: 0, initialSum: sum }
analyzeSum(root, result, sum)
return result.total
};
var analyzeSum = function(node, result, sum) {
if (!node) return
const extraSum = sum - node.val
if (extraSum === 0) {
result.total = result.total + 1
}
analyzeSum(node.left, result, extraSum)
analyzeSum(node.right, result, extraSum)
// handle it as root node.
analyzeSum(node.left, result, result.initialSum)
analyzeSum(node.right, result, result.initialSum)
}

submit 后, 卡在了如下测试用例中:

输入:
[1,null,2,null,3] // 该测试用例有点怪, 应该为 [1, null, 2, null, null, null, 3]
3
1
/ \
null 2
/ \
null 3
输出:
3
预期:
2

经过排查, analyzeSum 函数存在冗余的调用, 并未将根节点的递归分析总和的递归给解耦。调整如下:

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number}
*/
// recursive root node
var pathSum = function(root, sum) {
if (!root) return 0
const curTotal = analyzeSum(root, sum)
const leftChildTotal = pathSum(root.left, sum)
const rightChildTotal = pathSum(root.right, sum)
return curTotal + leftChildTotal + rightChildTotal
};
// recursive anlyze total value
var analyzeSum = function(node, expectSum) {
if (!node) return 0
const extraSum = expectSum - node.val
if (extraSum === 0) {
return 1 + analyzeSum(node.left, extraSum) + analyzeSum(node.right, extraSum)
} else {
return analyzeSum(node.left, extraSum) + analyzeSum(node.right, extraSum)
}
}